Integrand size = 67, antiderivative size = 26 \[ \int \frac {1}{16} e^{e^{\frac {1}{16} \left (80+18 x-e^5 x-x^2\right )}-x} \left (-16+e^{\frac {1}{16} \left (80+18 x-e^5 x-x^2\right )} \left (18-e^5-2 x\right )\right ) \, dx=e^{e^{5+x+\frac {1}{16} \left (2-e^5-x\right ) x}-x} \]
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Time = 0.17 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.08, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {12, 6838} \[ \int \frac {1}{16} e^{e^{\frac {1}{16} \left (80+18 x-e^5 x-x^2\right )}-x} \left (-16+e^{\frac {1}{16} \left (80+18 x-e^5 x-x^2\right )} \left (18-e^5-2 x\right )\right ) \, dx=e^{e^{\frac {1}{16} \left (-x^2-e^5 x+18 x+80\right )}-x} \]
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Rule 12
Rule 6838
Rubi steps \begin{align*} \text {integral}& = \frac {1}{16} \int e^{e^{\frac {1}{16} \left (80+18 x-e^5 x-x^2\right )}-x} \left (-16+e^{\frac {1}{16} \left (80+18 x-e^5 x-x^2\right )} \left (18-e^5-2 x\right )\right ) \, dx \\ & = e^{e^{\frac {1}{16} \left (80+18 x-e^5 x-x^2\right )}-x} \\ \end{align*}
Time = 1.18 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {1}{16} e^{e^{\frac {1}{16} \left (80+18 x-e^5 x-x^2\right )}-x} \left (-16+e^{\frac {1}{16} \left (80+18 x-e^5 x-x^2\right )} \left (18-e^5-2 x\right )\right ) \, dx=e^{e^{5-\frac {1}{16} x \left (-18+e^5+x\right )}-x} \]
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Time = 0.20 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85
method | result | size |
norman | \({\mathrm e}^{{\mathrm e}^{-\frac {x \,{\mathrm e}^{5}}{16}-\frac {x^{2}}{16}+\frac {9 x}{8}+5}-x}\) | \(22\) |
risch | \({\mathrm e}^{{\mathrm e}^{-\frac {x \,{\mathrm e}^{5}}{16}-\frac {x^{2}}{16}+\frac {9 x}{8}+5}-x}\) | \(22\) |
parallelrisch | \({\mathrm e}^{{\mathrm e}^{-\frac {x \,{\mathrm e}^{5}}{16}-\frac {x^{2}}{16}+\frac {9 x}{8}+5}-x}\) | \(22\) |
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Time = 0.25 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {1}{16} e^{e^{\frac {1}{16} \left (80+18 x-e^5 x-x^2\right )}-x} \left (-16+e^{\frac {1}{16} \left (80+18 x-e^5 x-x^2\right )} \left (18-e^5-2 x\right )\right ) \, dx=e^{\left (-x + e^{\left (-\frac {1}{16} \, x^{2} - \frac {1}{16} \, x e^{5} + \frac {9}{8} \, x + 5\right )}\right )} \]
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Time = 0.18 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {1}{16} e^{e^{\frac {1}{16} \left (80+18 x-e^5 x-x^2\right )}-x} \left (-16+e^{\frac {1}{16} \left (80+18 x-e^5 x-x^2\right )} \left (18-e^5-2 x\right )\right ) \, dx=e^{- x + e^{- \frac {x^{2}}{16} - \frac {x e^{5}}{16} + \frac {9 x}{8} + 5}} \]
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Time = 0.32 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81 \[ \int \frac {1}{16} e^{e^{\frac {1}{16} \left (80+18 x-e^5 x-x^2\right )}-x} \left (-16+e^{\frac {1}{16} \left (80+18 x-e^5 x-x^2\right )} \left (18-e^5-2 x\right )\right ) \, dx=e^{\left (-x + e^{\left (-\frac {1}{16} \, x^{2} - \frac {1}{16} \, x e^{5} + \frac {9}{8} \, x + 5\right )}\right )} \]
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\[ \int \frac {1}{16} e^{e^{\frac {1}{16} \left (80+18 x-e^5 x-x^2\right )}-x} \left (-16+e^{\frac {1}{16} \left (80+18 x-e^5 x-x^2\right )} \left (18-e^5-2 x\right )\right ) \, dx=\int { -\frac {1}{16} \, {\left ({\left (2 \, x + e^{5} - 18\right )} e^{\left (-\frac {1}{16} \, x^{2} - \frac {1}{16} \, x e^{5} + \frac {9}{8} \, x + 5\right )} + 16\right )} e^{\left (-x + e^{\left (-\frac {1}{16} \, x^{2} - \frac {1}{16} \, x e^{5} + \frac {9}{8} \, x + 5\right )}\right )} \,d x } \]
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Time = 0.32 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {1}{16} e^{e^{\frac {1}{16} \left (80+18 x-e^5 x-x^2\right )}-x} \left (-16+e^{\frac {1}{16} \left (80+18 x-e^5 x-x^2\right )} \left (18-e^5-2 x\right )\right ) \, dx={\mathrm {e}}^{-x}\,{\mathrm {e}}^{{\mathrm {e}}^{\frac {9\,x}{8}}\,{\mathrm {e}}^5\,{\mathrm {e}}^{-\frac {x^2}{16}}\,{\mathrm {e}}^{-\frac {x\,{\mathrm {e}}^5}{16}}} \]
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